Question

If ${\cot }^{2}x=\cot (x-y)\cot (x-z)$, then the value of $\cot 2x$ is

• A. $\frac{1}{2}(\tan y+\tan z)$
• B. $\frac{1}{2}(\cot y+\cot z)$
• C. $\frac{1}{2}(\sin y+\sin z)$
• D. $\frac{1}{2}(\cot y-\cot z)$

Answer 1

The correct option is B $\frac{1}{2}(\cot y+\cot z)$

We know that,
$\cot 2x=\frac{{\cot }^{2}x-1}{2\cot x}\cdots \left(1\right)$
Now, given equation,
${\cot }^{2}x=\cot (x-y)\cot (x-z)$
$\Rightarrow {\cot }^{2}x=\left(\frac{\cot x\cot y+1}{\cot y-\cot x}\right)\left(\frac{\cot x\cot z+1}{\cot z-\cot x}\right)\Rightarrow {\cot }^{2}x(\cot x-\cot y)(\cot x-\cot z)=(\cot x\cot y+1)(\cot x\cot z+1)\Rightarrow {\cot }^{2}x[{\cot }^{2}x-\cot x(\cot y+\cot z)+\cot y\cot z]=[{\cot }^{2}x\cot y\cot z+\cot x(\cot y+\cot z)+1]\Rightarrow {\cot }^{2}x[{\cot }^{2}x-\cot x(\cot y+\cot z\left)\right]=\left[\cot x\right(\cot y+\cot z)+1]\Rightarrow {\cot }^{4}x-1=\cot x(\cot y+\cot z)(1+{\cot }^{2}x)\Rightarrow ({\cot }^{2}x+1)({\cot }^2-1)=\cot x(\cot y+\cot z)(1+{\cot }^{2}x)\Rightarrow {\cot }^2-1=\cot x(\cot y+\cot z)(\because 1+{\cot }^{2}x\ne 0)\Rightarrow \frac{{\cot }^{2}x-1}{\cot x}=\cot y+\cot z$
From equation $\left(1\right)$,
$\therefore \cot 2x=\frac{1}{2}(\cot y+\cot z)$